Question

Xác định giá trị nhỏ nhất của biểu thức : T=(3x + 2y +1)^2 + (12x - my - 2)^2

Answer 1

- Ta thấy : \(\left\{{}\begin{matrix}\left(3x+2y+1\right)^2\ge0\forall x,y\\\left(12x-my-1\right)^2\ge0\forall x,y\end{matrix}\right.\)

=> \(\left(3x+2y+1\right)^2+\left(12x-my-2\right)^2\ge0\forall x,y\)

=> Min_T = 0 khi \(\left\{{}\begin{matrix}3x+2y+1=0\\12x-my-2=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}12x+8y+4=0\\12x-my-2=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}3x+2y+1=0\\8y+my+6=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}3x-\frac{12}{m+8}+1=0\\y=-\frac{6}{m+8}\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\frac{\frac{12}{m+8}-1}{3}\\y=-\frac{6}{m+8}\end{matrix}\right.\)

Vậy Min_T = 0 khi \(x=\frac{\frac{12}{m+8}-1}{3},y=-\frac{6}{m+8}\)